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I've been trying to solve the problem below for over an hour doing everything but writing it out by hand. I've tried looking for patterns in strings of lengths $2, 3, 4,$ and $5$ which have probabilities $1/4, 3/8, 8/16,$ and $19/32$, respectively, of having two or more consecutive $1$s.

I'm more interested in the formula and theory behind solving the problem than the answer itself. I've searched online for permutation and probability problems without success.

Problem:

For the purposes of this puzzle, a "string" means a sequences of zeros and ones, e.g. $1011$.

Consider all the different strings of length eight — a good first step would be to figure out how many of these exist.

If you pick a length-eight string uniformly at random, what's the probability that it contains two (or more) consecutive 1s?

Problem is from the app below: https://play.google.com/store/apps/details?id=atorch.statspuzzles

Linus Rastegar
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Howdoyoulikemyperm
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  • Hint: each position in the string can has two different 'choices' (either it's $0$ or $1$). So, for example, there are $2 \times 2 = 4$ choices for a two-digit string. Khan academy gives a decent introductory account here: https://www.khanacademy.org/math/precalculus/prob-comb – TheMathsGeek Apr 07 '17 at 14:14
  • Thank you. I know how to get the total number of permutations. 2^n where n is the length of the string. What I am having trouble with is the formula for calculating the number of permutations that will have two or more consecutive 1s. – Howdoyoulikemyperm Apr 07 '17 at 14:22
  • You can solve it by markov chains with the approach on my old answer: http://math.stackexchange.com/questions/2216650/getting-at-least-k-heads-in-a-row-l-times-over-after-at-most-m-flips/2218235#2218235 _ Just sum the last two states for ${\bf P}^8{\bf v}$ _ I can expand later if you want, but I'm in a hurry now. – mathreadler Apr 07 '17 at 14:54

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Let's try to solve the opposite problem first: counting the number of bitstrings with no consecutive ones.

Let $A_n$ be the set of bit strings of length $n$ with no consecutive $1$'s and let $a_n = |A_n|$, i.e., $a_n$ is the number of bit strings of length $n$ with no consecutive $1$'s. Clearly, $a_1 = 2, a_2 = 3$.

Consider a bit string in $A_n$. Then it can be formed in the following ways:

  1. $0$ followed by any bit string in $A_{n-1}$.
  2. $10$ followed by any bit string in $A_{n-2}$.

Thus, it follows that: $$a_n = a_{n-1} + a_{n-2}.$$

Using standard techniques for solving recurrence relations, you should be able to find a formula for $a_n$. If we let $b_n$ be the number of bit strings that have consecutive $1$'s, then since there are a total of $2^n$ bitstrings of length $n$, then $a_n+b_n = 2^n \implies b_n = 2^n-a_n$. Thus, the probability of getting a bitstring with consecutive $1$'s is:

$$\frac{2^n-a_n}{2^n} = 1-\frac{a_n}{2^n}.$$

benguin
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